What is the solution for y^{''}-4y^'=(x^2-3)sin(2x) ?

The solution for y^{''}-4y^'=(x^2-3)sin(2x) is

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y^{''}-4y^'=(x^2-3)sin(2x)

y^{^{''}} - 4 y^{^{'}} = (x^{2} - 3) sin (2 x)

y = c_{1} e^{4 x} + c_{2} + \frac{x ^{2} cos ( 2 x )}{10} - \frac{x ^{2} sin ( 2 x )}{20} - \frac{7 x sin ( 2 x )}{50} - \frac{x cos ( 2 x )}{50} + \frac{149}{1000} sin (2 x) - \frac{46}{125} cos (2 x)

Solution steps

y^{''} (x) - 4 y^{'} (x) = (x^{2} - 3) sin (2 x)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + \frac{x ^{2} cos ( 2 x )}{10} - \frac{x ^{2} sin ( 2 x )}{20} - \frac{7 x sin ( 2 x )}{50} - \frac{x cos ( 2 x )}{50} + \frac{149}{1000} sin (2 x) - \frac{46}{125} cos (2 x)

y = c_{1} e^{4 x} + c_{2} + \frac{x ^{2} cos ( 2 x )}{10} - \frac{x ^{2} sin ( 2 x )}{20} - \frac{7 x sin ( 2 x )}{50} - \frac{x cos ( 2 x )}{50} + \frac{149}{1000} sin (2 x) - \frac{46}{125} cos (2 x)

y^{^{''}} + 25 y = 2 sec (5 t)

2 y^{^{''}} + 3 y^{^{'}} + y = sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3

y^{^{''}} - 8 y^{^{'}} + 25 y = (5 x^{3} - 7) e^{- x}

x^{^{''}} + 4 x = cos (t)

What is the solution for y^{''}-4y^'=(x^2-3)sin(2x) ?
The solution for y^{''}-4y^'=(x^2-3)sin(2x) is